University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.3 - Area by Double Integration - Exercises - Page 772: 18



Work Step by Step

Our aim is to integrate the integral as follows: $ Area =\iint_D dA $ $\int_{0}^{2} \int^{0}_{x^2-4}dydx+\int_{0}^{{\sqrt{x}}} \space dy \space dx =\int_{0}^{2}(4-x^2)dx+\int_{0}^{4} \sqrt x \space dx $ or, $=[4x-\dfrac{1}{3}x^3]^2_0+[\dfrac{2x^{3/2}}{3}]^4_0$ or, $=(8-\dfrac{8}{3})+\dfrac{16}{3}$ and $ Area=\dfrac{32}{3}$
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